Question 183: Centre of Mass and Rotational Motion

Question

A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{	heta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It
bounces back and goes up to an angle $${{	heta _1}}$$. Then M is given by :

Accepted Answer

v = $$\sqrt {2g\ell \left( {1 - \cos {	heta _0}} ight)} $$
v
1
= $$\sqrt {2g\ell \left( {1 - \cos {	heta _1}} ight)} $$
By momentum conservation
m$$\sqrt {2gl\left( {1 - \cos {	heta _0}} ight)}  
$$
$$= M{V_m} - m\sqrt {2g\left( {1 - \cos 	heta } ight)} $$
$$ \Rightarrow $$$$m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {	heta _0}}  + \sqrt {1 - \cos {	heta _1}} } ight\}$$
$$ = $$ MV
m
and  e = 1 = $${{{V_m} + \sqrt {2g\ell \left( {1 - \cos {	heta _1}} ight)} } \over {\sqrt {2g\ell \left( {1 - \cos {	heta _0}} ight)} }}$$
$$\sqrt {2g\ell } $$ $$\left( {\sqrt {1 - \cos {	heta _0}}  - \sqrt {1 - \cos {	heta _1}} } ight) $$
$$= $$ V
m
. . .(I)
m$$\sqrt {2g\ell } \left( {\sqrt {1 - \cos {	heta _0}}  + \sqrt {1 - \cos {	heta _1}} } ight)$$
$$ = $$ MV
M
. . .(II)
Dividing
$${{\left( {\sqrt {1 - \cos {	heta _0}}  + \sqrt {1 - \cos {	heta _1}} } ight)} \over {\left( {\sqrt {1 - \cos {	heta _0}}  + \sqrt {1 - \cos {	heta _1}} } ight)}} = {M \over m}$$
By componendo divided
$${{m - M} \over {m + M}}$$ = $${{\sqrt {1 - \cos {	heta _1}} } \over {\sqrt {1 - \cos {	heta _0}} }} = {{\sin \left( {{{{	heta _1}} \over 2}} ight)} \over {\sin \left( {{{{	heta _0}} \over 2}} ight)}}$$
$$ \Rightarrow $$  $${M \over m} = {{{	heta _0} - {	heta _1}} \over {{	heta _0} + {	heta _1}}} \Rightarrow M = {{{	heta _0} - 	heta 1} \over {{	heta _0} + {	heta _1}}}$$

Answer

$${m \over 2}\left( {{{{	heta _0} + {	heta _1}} \over {{	heta _0} - {	heta _1}}}} ight)$$

Answer

$${m \over 2}\left( {{{{	heta _0} - {	heta _1}} \over {{	heta _0} + {	heta _1}}}} ight)$$

Answer

$$m\left( {{{{	heta _0} + {	heta _1}} \over {{	heta _0} - {	heta _1}}}} ight)$$

Answer

$$m\left( {{{{	heta _0} - {	heta _1}} \over {{	heta _0} + {	heta _1}}}} ight)$$

Centre of Mass and Rotational Motion

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